Let S be a set of four vectors in R4 with the property that every subset of S consisting of three vectors is linearly independent. Can we deduce that S itself is linearly independent? Justify your answer.

Currently studying for my linear algebra exam and have no idea where to begin on the question any help would be appreciated thanks


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    $\begingroup$ Try $(1,0,0,0),\,(0,1,0,0),\,(0,0,1,0),\,(1,1,1,0)$ $\endgroup$ – lulu Jul 19 '18 at 13:07
  • $\begingroup$ Oh do you have to take actual values $\endgroup$ – Molly Jul 19 '18 at 13:09
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    $\begingroup$ Certainly not. You should easily be able to use my example to build families of examples. But, of course, one example is enough to settle the point. $\endgroup$ – lulu Jul 19 '18 at 13:13

When you have no idea where to begin on the question, try to think about the closest problem you've encountered, try to think about something that strikes some similarity. If nothing comes to mind, and the question is essentially a yes/no question ( can we deduce? is the set linearly independent? etc.), take a guess. In this case - you could guess that the answer is "Yes, we can deduce." Then try to prove your guess. If you fail, maybe your guess was wrong; try the opposite guess.Look at some examples. You must have some examples somewhere. If you don't, and you are a Linear Algebra student, then your situation is not optimal, to put it mildly. In any case, getting someone else to answer your mathematical problems usually actually prevents you from learning anything. You never really own a solution you haven't come up with yourself. Yes, you can memorize it for a test, and possibly pass the test, but you will not know any more mathematics than someone who may have failed the test but the solutions they provided were their own.


Consider three vectors $v_1$, $v_2$ and $v_3$ that form a linearly independent set. Take $v_4=v_1+v_2+v_3$. Show that $\{v_1,v_2,v_4\}$ is linearly independent as well.

$$av_1+bv_2+cv_4=0$$we obtain$$(a+c)v_1+(b+c)v_2+cv_3=0$$so $c=0$ and consequently $a=b=0$.

By symmetry, also $\{v_2,v_3,v_4\}$ and $\{v_1,v_3,v_4\}$ are linearly independent.

Hence all three-element subsets of $\{v_1,v_2,v_3,v_4\}$ are linearly independent. However $\{v_1,v_2,v_3,v_4\}$ is linearly dependent.

Thus the situation is reproducible on every vector space of dimension $\ge3$.

  • $\begingroup$ I thought a set of vectors is linearly independent if and only if each of its subsets is linearly independent so would that not mean S was linear independent $\endgroup$ – Molly Jul 19 '18 at 13:51
  • $\begingroup$ @Molly This is not true, as you discovered here. Among the subsets you need also the whole set. For an infinite set $S$ (in an infinite dimensional space, of course), it is true that $S$ is linearly independent if and only if every finite subset is linearly independent. $\endgroup$ – egreg Jul 19 '18 at 13:56
  • $\begingroup$ Is this because v4 equals the sum of v1+v2+v3 $\endgroup$ – Molly Jul 19 '18 at 13:59
  • $\begingroup$ @Molly In this case it is so. But the example shows you can't conclude a set is linearly independent if every proper subset is linearly independent. $\endgroup$ – egreg Jul 19 '18 at 14:00
  • $\begingroup$ Is there any case where S would be linear independent $\endgroup$ – Molly Jul 19 '18 at 14:04

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